# A 100.0 mL sample of a solution that is 0.125 M in HClO and 0.130 M in NaClO has added to it 2.00 mL of 0.100 M HCl. The Ka for HClO is 2.9 x 10-8. What is the pH of the resulting solution?

## To start, I found the moles of each compound, then I added the moles of HClO and HCl to get my total moles of acid. Finally, I plugged it into the Henderson-Hasselbach equation. I found that 0.0002 moles of HCl was added to a solution that contained 0.0125 moles of HClO and 0.013 moles of NaClO So... -log(2.9 -log($2.9 \cdot {10}^{-} 8$) + log($\frac{0.013}{0.0127}$) pH=7.55 Does this seem right? I wasn't sure how to approach this problem and I am unsure about the answer I got. Thanks for any help!

Jul 7, 2018

$\text{pH} = 7.554$

#### Explanation:

Start by finding the quantity of each species in the $\text{NaClO}$ buffer immediately after the addition of $\text{HCl}$.

• $n \left(\text{HCl}\right) = c \cdot V = 2.00 \times {10}^{- 4} \textcolor{w h i t e}{l} m o l$
• $n \left(\text{HClO}\right) = c \cdot V = 1.25 \times {10}^{- 2} \textcolor{w h i t e}{l} m o l$
• $n \left(\text{NaClO}\right) = c \cdot V = 1.30 \times {10}^{- 2} \textcolor{w h i t e}{l} m o l$

Hydrochloric acid $\text{HCl}$ would react with $\text{NaClO}$ to produce a weak electrolyte $\text{HClO}$ through the following double decomposition reaction:

stackrel((color(grey)("strong")))("HCl"(aq)) + "NaClO" (aq) color(navy)(to) stackrel((color(grey)("weak")))("HClO"(aq)) + "NaCl"(aq)

$\text{NaClO}$ is in excess and therefore

$n \left(\text{HClO","final") = n("HClO","initial")+ n("HCl}\right)$
$\textcolor{w h i t e}{n \left(\text{HClO","final}\right)} = 1.2502 \times {10}^{- 2} \textcolor{w h i t e}{l} m o l$

$n \left(\text{NaClO", "final") = n("NaClO", "initial") - n("HCl}\right)$
$\textcolor{w h i t e}{n \left(\text{NaClO","final}\right)} = 1.2998 \times {10}^{- 2} \textcolor{w h i t e}{l} m o l$

The rest of $\text{NaClO}$ completely disassociates to produce ${\text{Na}}^{+}$ and ${\text{ClO}}^{-}$.

${\text{NaClO"(s) color(navy)(to) "Na"^(+)(aq) + "ClO}}^{-} \left(a q\right)$

$n \left(\text{ClO"^(-)) = n("NaClO}\right) = 1.30 \times {10}^{- 2} \textcolor{w h i t e}{l} m o l$

The hydrolysis of aqueous ${\text{ClO}}^{-}$ from $\text{NaClO}$ might pose an influence on the acidity (in other words, $\text{pH}$) of the solution. A "RICE" table can help with the derivation of the equilibrium proton concentration due to this reversible process.

R $\text{ ClO"^(-)(aq) + "H"_2"O"(l) color(darkgreen)(rightleftharpoons) "OH"^(-) (aq) + "HClO} \left(a q\right)$
I color(purple)(color(white)(ll)1.2998 xx 10^(-2) color(white)(-------) color(white)(ll)1.2502 xx 10^(-2)
C $\textcolor{\mathrm{da} r k g r e e n}{\text{ } - x \textcolor{w h i t e}{- - - - - - -} + x \textcolor{w h i t e}{- - -} + x}$
E $\textcolor{n a v y}{\textcolor{w h i t e}{l l} 1.2998 \times {10}^{- 2} \textcolor{w h i t e}{- - - - - - -} 1.2502 \times {10}^{- 2}}$
$\textcolor{n a v y}{\textcolor{w h i t e}{- -} - x \textcolor{w h i t e}{- - - - - l l l -} + x \textcolor{w h i t e}{- - -} + x}$
(all values in this table are in $m o l$)

${K}_{a} = 2.9 \times {10}^{- 8}$ for $\text{HClO}$ such that
${K}_{b} = \frac{1.0 \times {10}^{- 14}}{2.9 \times {10}^{- 8}} = 3.448 \times {10}^{- 7}$ for ${\text{ClO}}^{-}$, its conjugate base.

(n("OH"^(-)) * n("HClO")) / (n("ClO"^(-))) = (["OH"^(-)] * ["HClO"]) / (["ClO"^(-)]) * V= K_b * V = 3.4 xx 10^(-7) * 0.102

Where $n \left({\text{OH}}^{-}\right)$, $n \left(\text{HClO}\right)$, and $n \left({\text{ClO}}^{-}\right)$ the $\textcolor{n a v y}{\text{quantities}}$ (in number of moles) of each of the respective species present in the solution at its $\textcolor{n a v y}{\text{equilibrium}}$ position. With reference to the RICE table,

((x) * (1.2502 xx 10^(-2) + x)) / (1.2998 xx 10^(-2) - x) = 3.448 xx 10^(-7) * 0.102

Solving this equation for $x$ yields $x = 3.6 \textcolor{g r e y}{56} \times {10}^{- 8} \textcolor{w h i t e}{l} \textcolor{p u r p \le}{m o l}$. Therefore $n \left({\text{OH}}^{-}\right) = x = 3.6 \textcolor{g r e y}{56} \times {10}^{- 8} \textcolor{w h i t e}{l} \textcolor{p u r p \le}{m o l}$ of ${\text{OH}}^{-}$ would eventually be present in the solution as it comes to its equilibrium state.

The final solution has a volume of $100.0 + 2.00 = 102.0 \textcolor{w h i t e}{l} m l$ (assuming no volume change due to the mixing.) Thus $\left[{\text{OH}}^{-}\right] = \frac{n}{V} = 3.584 \times {10}^{- 7} \textcolor{w h i t e}{l} m o l$ at equilibrium.

"pOH" = -log["OH"^(-)] = 6.445
and therefore
$\text{pH" = "pKw" - "pOH} = 7.554$.